Swapping Two Variables

Swapping the values of two variables is a common task in programming, for example when sorting. In main, without a function, it looks easy at first:

int a = 3;
int b = 8;
a = b;
b = a;
printf("%d %d\n", a, b);

This prints:

8 8

The line a = b; overwrote the original value of a, the 3, and it was lost. It's like swapping the drinks in two full glasses: you need a third, empty glass. In a program, that glass is a helper variable, where you first store the original value of a:

int a = 3;
int b = 8;
int temp = a;
a = b;
b = temp;
printf("%d %d\n", a, b);

This prints:

8 3

A function swap has to change two variables of its caller, so both parameters must be pointers: void swap(int *a, int *b). Inside, it does the same three steps as above, only with *a and *b instead of a and b. The helper variable temp stays an ordinary int, because it holds a value, not an address. The call passes two addresses: swap(&x, &y);.

Watch out for these mistakes:

• When you forget the stars (int *temp = a; a = b; b = temp;), the function swaps the two pointers, its own copies of the addresses. The program compiles and runs, but the variables in main stay as they were.
• A function that already has pointers passes them on without &. Inside a function with the parameters int *a and int *b, you call swap(a, b), because a and b already are addresses. swap(&a, &b) ends with the error incompatible pointer types passing 'int **' to parameter of type 'int *'; remove &. int ** is the address of a pointer, which you don't need here, and the message tells you what to do: remove the &.